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samnatha

in a particular game a player throws two fair dice together. if their total is 7 or 11 he wins if their total is 2,3 or 12 he looses for any other total, he neither wins or looses. (i) find the probability of winning (ii) find the probability of losing

  • 11 months ago
  • 11 months ago

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  1. tcarroll010
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    Start with listing all your possible dice pair outcomes and count the number of 7's and 11's out of all possible outcomes. That gives your probability of winning. Same process with the losing for 2, 3, and 12.

    • 11 months ago
  2. tcarroll010
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    So start with "all possible outcomes". How many of those are there? hint: each die has 6 faces.

    • 11 months ago
  3. samnatha
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    am 2 dice so 12 ?

    • 11 months ago
  4. tcarroll010
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    Better start with one die. How many outcomes?

    • 11 months ago
  5. samnatha
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    6 ?

    • 11 months ago
  6. tcarroll010
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    Yes. Now for EACH GIVEN outcome on that first die, you can have 6 outcomes on the second die. So, 6 x 6 = 36 possible outcomes for a roll of 2 die. Now, it really would be beneficial for you to write out on paper a 2-dimensional table where horizontal is one die and vertical is the other die and make a table of 36 rolls. That will benefit you immensely here.

    • 11 months ago
  7. samnatha
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    is the answer for part (i) 2over 9 ?

    • 11 months ago
  8. tcarroll010
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    Yes, 2/9 for winning. That comes from 6 7's and 2 11's inning rolls out of a possible 36: 8/36 = 2/9 Good job!

    • 11 months ago
  9. samnatha
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    yeah ok thank you is is the same method for loosing aswell ?

    • 11 months ago
  10. tcarroll010
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    Very similar. Add up all 2's (there's just 1), all 3's (there are 2) and all 12's (there's 1) for that total over 36.

    • 11 months ago
  11. samnatha
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    1 over 9 ?

    • 11 months ago
  12. tcarroll010
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    Yes, because 4/36 = 1/9. This would be a nice game to play because you win more than lose!

    • 11 months ago
  13. samnatha
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    thank you :) i actuallt have a similar question to that and i was wondering if i got it right or not do you mind checking for me ?

    • 11 months ago
  14. tcarroll010
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    Good luck in all of your studies and thx for the recognition! @samnatha I'll check it if you like.

    • 11 months ago
  15. samnatha
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    thank you i think i sent to you in mail :)

    • 11 months ago
  16. samnatha
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    i'll just put it here aswell :)

    • 11 months ago
  17. samnatha
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    this circle is divided into 6 equal sectors. you pay €10 to spin the arrow and you win the amount in the sector where the arrow stops. what is the expected amount you win or loose in this game ? theres a picture of a cirlce aswell divided into 6 sections in each section there is different numbers which are €0 €30 €10 €10 €0 and €5

    • 11 months ago
  18. tcarroll010
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    Since you pay 10 to play, we are really talking about the "net" winning or losing, so you can actually re-do the 6 sectors as a "net" amount: -10, +20, 0, 0, -10, -5 You add those up and divide by 6 to get -5/6. Not a good game to play. Over time, you lose big.

    • 11 months ago
  19. samnatha
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    ok i'm slighlt confused on how you did all that not great at probability

    • 11 months ago
  20. tcarroll010
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    np. Each sector is equally likely, and for each sector, we can conceptually see ourselves handing over 10 (at first, to play) and then getting back some money (if any for that sector). So, for example, if we landed on "get 30", it's a net of 20 because we had to pay 10 up front. Your expected value is the net amount times the probability of landing in that sector. It is 1/6 times each sector and then all added up. Or you could just add them all up first: (1/6)(-10) + (1/6)(20) + (1/6)(0) + (1/6)(0) + (1/6)(-10) + (1/6)(-5) or what I did: (1/6)(-10 + 20 + 0 + 0 + -10 -5)

    • 11 months ago
  21. samnatha
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    ooh ok thank you i am use to doing it the other way so that made more sense thanks alot :)

    • 11 months ago
  22. tcarroll010
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    yw!

    • 11 months ago
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